Everyone likes to prove that Dirichlet convolution is a group operation on the multiplicative arithmetic functions, but what consequence does this have?

Does any important theorem use this fact?

Can general group theory lead to results about these functions (or even better, about numbers) from the this theorem?

Furthermore there are two ring structures on this set, the usual pointwise ring as well as the ring with convolution.

I would like to extend the same question for these.

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    $\begingroup$ If I can piggyback onto this question: a couple of years ago, I learned the fact (from a book on probabilistic number theory!) that the ring of all $\mathbb{C}$-valued arithmetic functions with pointwise addition and convolution product is a UFD. Ever since I've been looking for some nice application of this curious fact. Does anyone have one? $\endgroup$ – Pete L. Clark Feb 9 '11 at 0:35
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    $\begingroup$ It's extra structure, and experience has taught mathematicians that we should always look for extra structure in order to exploit it. $\endgroup$ – Qiaochu Yuan Feb 9 '11 at 0:35
  • $\begingroup$ @Pete: Based on google search, I guess that book is Tenenbaum's Introduction to analytic and probabilistic number theory, where on p.26 the author writes: These operations give the set A of arithmetic functions the structure of a commutative ring with unity isomorphic to that of formal Dirichlet series. Cashwell & Everett (1959) showed that this ring is factorial, that is to sayan integral domain whose quotient by the group of units satisfies the fundamental theorem of arithmetic. $\endgroup$ – Martin Sleziak Nov 26 '11 at 13:04
  • $\begingroup$ @Martin: yes, that's the book. $\endgroup$ – Pete L. Clark Nov 26 '11 at 17:17

James Delaney's paper "Groups of Arithmetical Functions" (Mathematics Magazine 78 (2), 83-97, 2005) may be worth a look. Here are some of his main results.

The group $U$ of units of the ring of arithmetic functions (pointwise addition and Dirichlet convolution ($\ast$) as multiplication) can be expressed as the direct sum $U = C \oplus U_M \oplus U_A$, where $C$ is the subgroup of scalar functions, $U_M$ is the subgroup of multiplicative functions, and $U_A$ is the subgroup of what he calls "anti-multiplicative" functions defined by $f(1) = 1$ and $f(p^k) = 0$ when $p^k$ is a prime power with $k > 0$.

He then shows, if $U_1 = U_M \oplus U_A$, then $(U_1, \ast)$ is a divisible torsion-free group and thus can be viewed as a vector space over the rationals. Therefore $U_M$ and $U_A$ are divisible subgroups and are thus complementary subspaces when $U_1$ is regarded as a vector space. This also implies that the $n$th root of a multiplicative function is multiplicative.

Finally, he proves that the functions $\{\epsilon_{\alpha} \lambda_{\beta} | \alpha, \beta \in \mathbb{C}, \beta \neq 0\}$ are linearly independent, where $\epsilon_{\alpha}(n) = n^{\alpha}$ and $\lambda_{\beta}(p_1^{k_1} \cdots p_r^{k_r}) = \beta^{k_1 + \cdots + k_r}$ (a generalization of Liouville's $\lambda$).

I have pretty much paraphrased straight from Delaney's paper. My algebra background is too deficient for me to be able to comment on how interesting (or uninteresting) these results actually are or how well they satisfy what the OP is looking for. (Perhaps someone else could help out with that?)

  • $\begingroup$ It's late here, and I'll deal with any comments that require edits to the answer (or even deleting the answer if it's not helpful at all) in the morning. $\endgroup$ – Mike Spivey Feb 9 '11 at 7:42
  • $\begingroup$ please don't delete this answer! I hadn't seen Delaney's paper before, and it seems rather nice. Especially, I can imagine showing it to an honors undergraduate or master's student who wants a subject to do research on which does not require a tremendous amount of background reading. $\endgroup$ – Pete L. Clark Feb 9 '11 at 10:19
  • $\begingroup$ @Pete: Thanks for the confirmation that this was a valid answer to the OP's question. (Woo-hoo! My first contribution to the site in the area of algebra! :) ) $\endgroup$ – Mike Spivey Feb 9 '11 at 17:12

Because there is a group homomorphism into Dirichlet series.

This can be used to relate zeta values e.g. since $\varphi \star 1 = I$ leads to $\sum \varphi(n)/n^s = \zeta(s-1)/\zeta(s)$.

It can probably be used to prove arithmetic statements using zeta functions but I don't have any examples of that.


I saw a talk by John Thompson, not that I knew what it was about, looking at a copy of SL(2,Z) in Dirichlet series under convolution. This paper http://arxiv.org/abs/0803.1121 might be something...


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